1/[x^2.(x+2)^2]â¡ A/x+B/x^2 +C/(x+2)+D/(x+2)^2
=>
1â¡ Ax(x+2)^2+B(x+2)^2 +Cx^2.(x+2)+Dx^2
x=0, =>B=1/4
x=-2, =>D=1/4
coef. of x
4A+4B=0
4A+1 =0
A= -1/4
coef. of x^3
A+C=0
C = 1/4
1/[x^2.(x+2)^2]â¡ (1/4) [ -1/x+1/x^2 +1/(x+2)+1/(x+2)^2]
â«dx/[x^2.(x+2)^2]
=(1/4)â«[ -1/x+1/x^2 +1/(x+2)+1/(x+2)^2] dx
=(1/4)(ln|(x+2)/x| - 1/x - 1/(x+2) ) + C
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